{"id":89777,"date":"2021-07-19T12:25:44","date_gmt":"2021-07-19T06:55:44","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=89777"},"modified":"2021-07-19T14:30:50","modified_gmt":"2021-07-19T09:00:50","slug":"1st-puc-biology-model-question-paper-9","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-biology-model-question-paper-9\/","title":{"rendered":"1st PUC Biology Model Question Paper 9 with Answers"},"content":{"rendered":"

Students can Download 1st PUC Biology Model Question Paper 9 with Answers, Karnataka 1st PUC Biology Model Question Papers with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Biology Model Question Paper 9 with Answers<\/h2>\n

Time: 3.15 Hours
\nMax Marks: 70<\/p>\n

General Instructions:<\/p>\n

    \n
  1. The question paper consists of four parts A, B, C, and D.<\/li>\n
  2. All the parts are compulsory.<\/li>\n
  3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.<\/li>\n<\/ol>\n

    Part – A<\/span><\/p>\n

    Answer the following questions in one word \/ one sentence each: ( 10 \u00d7 1 = 10 )<\/span><\/p>\n

    Question 1.
    \nWhat is a herbarium?
    \nAnswer:
    \nA herbarium is defined as a collection of plants, that have been dried, pressed and preserved on sheets.<\/p>\n

    Question 2.
    \nDefine venation.
    \nAnswer:
    \nThe mode of arrangement of veins and veinlets is called venation.<\/p>\n

    Question 3.
    \n\u201cEarthworms are called friends of farmers\u201d. Justify the statement.
    \nAnswer:
    \nAs they increase soil fertility by their worm castings and by making the soil porous by burrowing.<\/p>\n

    Question 4.
    \nWhat are chromoplasts?
    \nAnswer:
    \nPlastids contain coloured pigments except for green pigment chlorophyll.<\/p>\n

    Question 5.
    \nName the organic compounds (non-protein constituents) present in certain enzymes that are tightly bound to the apoenzyme.
    \nAnswer:
    \nProsthetic group.<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nWhy are mitochondria called \u201cpowerhouses of the cell\u201d?
    \nAnswer:
    \nProduce cellular energy in the form of ATP.<\/p>\n

    Question 7.
    \nDefine osmosis.
    \nAnswer:
    \nIt is a special type of diffusion in which only the solvent or water molecules move from a region of their higher concentration to a region of their lower concentration through a semi-permeable membrane.<\/p>\n

    Question 8.
    \nName the essential element that is present in chlorophyll.
    \nAnswer:
    \nMagnesium.<\/p>\n

    Question 9.
    \nName the red coloured oxygen storing pigment present in muscles.
    \nAnswer:
    \nMyoglobin.<\/p>\n

    Question 10.
    \nMention the hormone that regulates the 24-hour (diurnal) rhythm of the human body.
    \nAnswer:
    \nMelatonin.<\/p>\n

    PART-B<\/span><\/p>\n

    Answer any FIVE of the following questions in 3-5 sentences each, wherever applicable: (5 \u00d7 2 = 10)<\/span><\/p>\n

    Question 11.
    \nMention two rules of binomial nomenclature.
    \nAnswer:<\/p>\n

      \n
    1. Scientific names are, generally, written in Latin or derived from Latin, irrespective of their origin.<\/li>\n
    2. The scientific names are written in italics or underlined. The first word denotes the name of the genus, and the second word denotes the species.<\/li>\n
    3. The generic name starts with a capital letter, while the specific name starts with a small letter.<\/li>\n<\/ol>\n

      Question 12.
      \nWrite any two classes of Kingdom Fungi with one example for each class.
      \nAnswer:
      \nClass: Zygomycetes eg., Rhizopus
      \nClass: Ascomycetes eg., yeast.<\/p>\n

      Question 13.
      \n(a) Why bryophytes are called amphibians of the plant kingdom?
      \n(b) Cycas is considered a naked seeded plant. Give a scientific reason.
      \nAnswer:
      \n(a) Bryophytes can live in soil but require water to complete sexual reproduction.
      \n(b) Ovules are not enclosed by my ovary wall and remain exposed.<\/p>\n

      Question 14.
      \nWrite two differences between cartilaginous and bony fishes.
      \nAnswer:<\/p>\n\n\n\n\n\n\n\n
      Chondrichthyes<\/strong><\/td>\nOsteichthyes<\/strong><\/td>\n<\/tr>\n
      1. Skeleton is made up of cartilage.<\/td>\n1. Skeleton is made up of bones.<\/td>\n<\/tr>\n
      2. Exclusively marine.<\/td>\n2. Both marine and freshwater forms are seen.<\/td>\n<\/tr>\n
      3. They possess placoid scales.<\/td>\n3. Possess cycloid or ctenoid scales.<\/td>\n<\/tr>\n
      4. Mouth and nostrils are situated on the ventral side of the body.<\/td>\n4. Mouth and nostril are terminal in their position.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      Question 15.
      \nList two functions of Golgi bodies.
      \nAnswer:<\/p>\n

        \n
      1. It secretes various enzymes and hormones.<\/li>\n
      2. It forms secretory vesicles which store enzymes, proteins, carbohydrates etc.<\/li>\n
      3. It takes part in the formation of the cell plates and cell walls along with ER at the time of telophase in plant cells.<\/li>\n
      4. It helps in the formation of primary Lysosomes.<\/li>\n<\/ol>\n

        Question 16.
        \nDefine respiratory quotient, Write the RQ value for glucose.
        \nAnswer:
        \nIt is defined as the ratio of the volume of CO2<\/sub> evolved to the volume of O2<\/sub> utilized.
        \nRQ = \\(\\frac{\\text { Volume of } \\mathrm{CO}_{2} \\text { released }}{\\text { Volume of } \\mathrm{O}_{2} \\text { utilised }}\\)
        \nRQ value for carbohydrates is one.<\/p>\n

        Question 17.
        \nDifferentiate long-day plants and short-day plants.
        \nAnswer:<\/p>\n

          \n
        1. Long Day Plants: These flower in photoperiod more than critical day length, eg: Wheat, oats etc.<\/li>\n
        2. Short Day Plants: These flower in photoperiod less than critical day length. e.g: Tobacco, Chrysanthemums etc.<\/li>\n<\/ol>\n

          \"KSEEB<\/p>\n

          Question 18.
          \nMention two types of movements exhibited by the cells of the human body with an example for each.
          \nAnswer:
          \n1. Amoeboid Movement\/Pseudopodial movement :
          \nMacrophages and leucocytes exhibit amoeboid movement. It is also seen in the amoeba. Amoeboid movement is affected by pseudopodia formation by cytoplasmic streaming. It also involves cytoskeletal elements like microfilaments.<\/p>\n

          2. Ciliary Movement:<\/p>\n

            \n
          • These movements are caused by fine versatile hair called cilia.<\/li>\n
          • Ciliary movement occurs in those hollow tubular visceral organs, that is internally lined by ciliated epithelium.<\/li>\n
          • Ciliary movement in the trachea helps in removing foreign substances or particles.<\/li>\n
          • Ciliary movement in the fallopian tube causes the ova to move to the uterus.<\/li>\n<\/ul>\n

            PART-C<\/span><\/p>\n

            Answer any FIVE of the following questions in 40-80 words each, wherever applicable: (5 \u00d7 3 = 15)<\/span><\/p>\n

            Question 19.
            \nSchematically represent the Haplontic life cycle in plants.
            \nAnswer:
            \nHaplontic life cycle
            \n\"1st<\/p>\n

              \n
            1. The dominant phase in the life cycle is a free-living gametophyte (haploid).<\/li>\n
            2. The diploid phase is represented only by the single-celled zygote.<\/li>\n
            3. The zygote undergoes meiosis (zygotic meiosis) and forms haploid spores which germinate to form the gametophyte. e.g Spimgpra Chiamydomonas etc.<\/li>\n<\/ol>\n

              Question 20.
              \nExplain any three types of aestivation in lowering plants.
              \nAnswer:
              \nAestivation of calyx and corolla:
              \nThe mode of arrangement of sepals, petals or even tepals in a flower bud is called aestivation. The different kinds of aestivation are as follows:<\/p>\n

                \n
              1. Valvate aestivation: When sepals, petals or tepals are not overlapping.<\/li>\n
              2. Imbricate aestivation: When out of the total number of sepals, petals or tepals, one is completely out, one is completely in and the rest are in and out.<\/li>\n
              3. Descending imbricate aestivation: When the standard petals are large and overlap the two wing petals, which in turn overlap the keel petals. It \u00a1s technically known as vexillary aestivation.
                \nNote: It is characteristic of the members of the subfamily Papilionoideae (Papilionaceae).
                \ne.g: Pea, Bean, Indigofera, Tephrosia etc.<\/li>\n
              4. Ascending imbricate aestivation – When the small standard petal is completely \u00a1n and is overlapped by the lateral wing petals which in turn are overlapped, by the keel petals.
                \nNote: it is characteristic of the sub-family Caesalpinioideae (Caesalpinae).
                \ne.g: Caesalpinia puicherhima, Delonix regia, etc.<\/li>\n
              5. Quincuncial aestivation – When out of the total number of sepals or petals, two are completely out, two are completely in and the rest are in and out.<\/li>\n
              6. Contorted or twisted aestivation – When all sepals or petals are both in and out.<\/li>\n<\/ol>\n

                Question 21.
                \nDifferentiate homopolymers and heteropolymers (polysaccharides) with an example each.
                \nAnswer:
                \nHomopolymers are made up of only one type of monosaccharide, eg., cellulose, starch. Heteropolymers are made up of different types of monosaccharides, eg., chitin.<\/p>\n

                \"KSEEB<\/p>\n

                Question 22.
                \nWrite the schematic representation of steps involved in the formation of ethyl alcohol during fermentation of glucose.
                \nAnswer:
                \nThe first step of anaerobic respiration is glycolysis, in which glucose is incompletely oxidised to 2 pyruvic acid molecules with 2 ATP and 2NADH2 formed as by-products.<\/p>\n

                Pyruvic acid formed is used for alcoholic fermentation or lactic acid fermentation.<\/p>\n

                In alcoholic fermentation, the pyruvic acid is first decarboxylated to acetaldehyde liberating CO2<\/sub>. Later acetaldehyde is reduced to ethyl alcohol by NADH2<\/sub> of glycolysis.<\/p>\n

                This process occurs in anaerobes like yeast and bacteria.<\/p>\n

                In lactic acid fermentation, the pyruvic acid is directly reduced to lactic acid by NADH2<\/sub> of glycolysis. NO CO2<\/sub> is liberated. It occurs in lactobacilli and skeletal muscles.
                \n\"1st<\/p>\n

                Question 23.
                \nWrite about any three disorders of the digestive system.
                \nAnswer:
                \nJAUNDICE IICTERUSI:
                \nIt is the most common digestive disorder. Jaundice relates to the yellow appearance of skin, urine, nails and cornea of the eye due to an increased Bilirubin pigment (Bile pigment).<\/p>\n

                Jaundice is of three types:
                \n1. Haemolytic Jaundice or Neonatal Jaundice or Pre hepatic Jaundice: It is commonly found in newly born babies up to one week of age. It is due to excessive haemolysis (Destruction of RBC\u2019s).<\/p>\n

                2. Infective or Hepatocellular Jaundice: It is due to the infection of the hepatitis virus of the liver. So, the liver becomes unable to process Bilirubin. It results in a heavy accumulation of Bilirubin which leads to Hepatocellular Jaundice.<\/p>\n

                3. Obstructive Jaundice or Extra Hepatic jaundice: It occurs due to the obstruction in the Bile duct. The obstruction of the Bile duct is due to gallstone formation. The bile duct is thus blocked, and bile juice cannot enter into the duodenum.<\/p>\n

                Causes:<\/p>\n

                  \n
                1. Liver damage due to viral infection, and consumption of alcohol.<\/li>\n
                2. Increased Bilirubin pigment in the body.<\/li>\n
                3. Excessive haemolysis.<\/li>\n
                4. Blockage of bilk duct due to gallstone.<\/li>\n
                5. Presence of tumours in the bile duct.<\/li>\n<\/ol>\n

                  Effects:<\/p>\n

                    \n
                  1. Yellow colouration of the skin, eyes and urine.<\/li>\n
                  2. Fever, loss of appetite, nausea, vomiting and headache.<\/li>\n
                  3. Infective jaundice leads to bleeding, and shrinkage of the liver [Cirrhosis]<\/li>\n
                  4. Unconsciousness and finally it may lead to death.<\/li>\n<\/ol>\n

                    Prevention:
                    \nIt can be prevented by avoiding the consumption of alcohol, and also by avoiding the consumption of contaminated food and water.<\/p>\n

                    Treatment :<\/p>\n

                      \n
                    1. Vaccines and anti-viral drugs are used against infective jaundice.<\/li>\n
                    2. Obstructive jaundice can be treated by surgery.<\/li>\n<\/ol>\n

                      Question 24.
                      \nWhat is: (a) Tidal volume, (b) Residual volume, (c) Vital capacity?
                      \nAnswer:
                      \n(a) Tidal volume (VT): It is the volume of air inhaled or exhaled with each inhalation and exhalation without any extra effort. Its volume \u00a1s 500 ml.<\/p>\n

                      (b) Residual volume (RV): It is the volume of air that remains inside the lungs at the end of forceful expiration. Its volume is 1100 – 1200 ml.<\/p>\n

                      (c) Vital Capacity (VC): It is the total volume of air that can be expelled from the Lungs during maximum exhalation.
                      \nVC = TV + IRV + ERV
                      \n= 500 + 2500 + 1100
                      \n= 4100 ml approx.<\/p>\n

                      \"KSEEB<\/p>\n

                      Question 25.
                      \nHow are differentiation, dedifferentiation and redifferentiation different from each other?
                      \nAnswer:
                      \nDifferentiation: It is the conversion (maturation) of meristematic cells, to regain the power to divide under certain conditions.<\/p>\n

                      Dedifferentiation: Permanent cells like parenchyma can revert back to meristematic activities (cell division) to form secondary persistent. This conversion is called Dedifferentiation.<\/p>\n

                      Redifferentiation: If the secondary meristem divides to produce new cells, which further differentiate to form micelle cells, and once again lose the capacity to divide, then it is called Redifferentiation.<\/p>\n

                      Question 26.
                      \nClassify the animals with an example each based on the chief excretory product produced in them.
                      \nAnswer:
                      \nBased on the type of excretory product produced, animals are divided into 3 groups:
                      \n1. Anunonoteic animals: Animals that excrete ammonia as nitrogenous waste products are called ammonotelic animals.
                      \ne.g: Aquatic invertebrates and bony fish, larvae of frog (Tadpole).<\/p>\n

                      2. Ureotelic animals: Animals that excrete urea are called ureotelic animals. Urea is produced by the liver during the ornithine cycle. Urea is stored in the urinary bladder in a dissolved state called urine.
                      \ne.g: Cartilaginous fish (Shark), Mammals and Amphibians.<\/p>\n

                      3. Uricotelic animals: Animals that excrete uric acid as nitrogenous waste products are called uricotelic animals.
                      \ne.g: Reptiles, birds.<\/p>\n

                      PART-D (Section-I)<\/span><\/p>\n

                      Answer any FOUR of the following questions in 200-250 words each, wherever applicable: (4 \u00d7 5 = 20)<\/span><\/p>\n

                      Question 27.
                      \nList five characteristic features of birds.
                      \nAnswer:<\/p>\n

                        \n
                      1. Birds are homeothermic (warm-blooded) animals.<\/li>\n
                      2. The body is boat-shaped and is adapted in such a way as to offer the least resistance while flying.<\/li>\n
                      3. The body is differentiated into an ahead, long neck, a trunk and a short tail.<\/li>\n
                      4. The body is covered by feathers, which provide insulation.<\/li>\n
                      5. Forelimbs are modified into wings for flight. Wings are provided with feathers, which are bad conductors of heat.<\/li>\n
                      6. The hind limbs are shifted forwards to balance the body while walking. They are variously modified for perching, walking, running, swimming etc.,<\/li>\n
                      7. Hind limbs have 4 toes and are covered, with epidermal scales.<\/li>\n
                      8. The endoskeleton is fully ossified (bony). Bones are tubular and hollow fond have air cavities or air sacs (pneumatic bones).<\/li>\n
                      9. The bones of the skull are fused.<\/li>\n
                      10. Jaws are devoid of teeth but covered by horny sheaths and are modified to form a beak.<\/li>\n
                      11. The beak is variously modified to suit individual food habits.<\/li>\n<\/ol>\n

                        \"KSEEB<\/p>\n

                        Question 28.
                        \nDescribe the important events that occur during anaphase and telophase of mitosis.
                        \nAnswer:
                        \nAnaphase (Gr:ana-up; phases-stage):
                        \nAnaphase is characterized by the following features.<\/p>\n

                          \n
                        • The two chromatids of each chromosome separate comp leteb\u00e7 to become daughter chromosomes. The two daughter chromosomes move away from the equator towards the opposite poles.<\/li>\n
                        • During the anaphasic movement of chromosomes, the centromeres lead the path and the arms trail behind. As a result, the anaphasic chromosomes appear V, L, J and L shaped.<\/li>\n<\/ul>\n

                          Telophase (Gr:telos-end; phases-stage):
                          \nIt is characterized by the following features.<\/p>\n

                            \n
                          • Here, reversal of the prophasic events occur. The daughter chromosomes move and reach the opposite poles where they become thin and thread-like again. These threads overlap one another to form a fine chromatin network.<\/li>\n
                          • The spindle fibres disintegrate and disappear.<\/li>\n
                          • Reconstitution of a new nuclear envelope occurs. Nucleolus reappear.<\/li>\n<\/ul>\n

                            Question 29.
                            \nDraw a labelled diagram showing the anatomy of the leaf of a typical monocot plant.
                            \nAnswer:
                            \n\"1st<\/p>\n

                            Question 30.
                            \n(a) Carbon, nitrogen, phosphorous, sulphur, calcium, etc, are considered essential elements. What are the criteria for considering them as essential? (3M)
                            \n(b) What is the role of leg-haemoglobin in leguminous plants? (1M)
                            \n(c) Name a soil bacterium that helps in nitrification. (1M)
                            \nAnswer:
                            \n(a) The element must be absolutely necessary for supporting normal growth and reproduction, and in the absence of the element, the plants do not complete their life cycle.<\/p>\n

                              \n
                            • The requirement of the element must be specific and not replaceable by another element, i.e. deficiency of any one element cannot be met by supplying some other element.<\/li>\n
                            • The element must be directly involved in the metabolism of the plant.<\/li>\n<\/ul>\n

                              (b) Leg haemoglobin protects nitrogenase enzyme from oxygen and acts as an oxygen scavenger.
                              \n(c) Nitrosomonas.<\/p>\n

                              Question 31.
                              \nWrite the schematic representation of the Calvin cycle.
                              \nAnswer:
                              \nIt occurs \u00a1n grana of the chloroplast.
                              \nIt utilises the assimilatory powers i.e., ATP and NADPH2 produced \u00a1n light reaction.
                              \n\"1st
                              \nSteps:<\/p>\n

                                \n
                              1. Phosphorylation<\/li>\n
                              2. Carbon dioxide fixation<\/li>\n
                              3. Phosphorylation and reduction.<\/li>\n
                              4. Regeneration and RUMP<\/li>\n
                              5. Glucose formation.<\/li>\n<\/ol>\n

                                Note:- To complete a Calvin cycle, 18 ATPs and 12 NADPH2 molecules are required.<\/p>\n

                                Question 32.
                                \nMention one function of each of the following hormones:
                                \n(a) ADH, (b) Thymosin, (c) Glucagon, (d) Atrial natruiuretic fact, (e) Erythropoietin.
                                \nAnswer:
                                \n(a) ADH: Stimulates reabsorption of water and electrolytes in the kidney tubules.
                                \n(b) Thymosin: Promotes the differentiation of T – lymphocytes to provide cell-mediated immunity and promotes the production of antibodies to provide humoral immunity.
                                \n(c) Glucagon: Promotes glycogenolysis and increases blood sugar level.
                                \n(d) ANF: Causes dilation of the blood vessels and reduces blood pressures.
                                \n(e) Erythropoietin : stimulates erythropoiesis \/ formation of RBCS.<\/p>\n

                                Section-II<\/span><\/p>\n

                                Answer any THREE of the following questions in 200-250 words each, wherever applicable: (3 \u00d7 5 = 15)<\/span><\/p>\n

                                Question 33.
                                \nDraw a labelled diagram showing the ultrastructure of a plant cell.
                                \nAnswer:
                                \n\"1st
                                \nPlant Cell – Ultrastructure<\/p>\n

                                \"KSEEB<\/p>\n

                                Question 34.
                                \nClassify simple epithelium and mention the structural modifications of cells in them.
                                \nAnswer:
                                \nSimple epithelium: Simple epithelium is further classified into:<\/p>\n

                                  \n
                                1. Squamous epithelium.<\/li>\n
                                2. Columnar epithelium.<\/li>\n
                                3. Cuboidal epithelium.<\/li>\n
                                4. Ciliated epithelium.<\/li>\n
                                5. Glandular epithelium.<\/li>\n<\/ol>\n

                                  1. Squamous epithelium :
                                  \n1. It occurs in the skin of vertebrates, alveoli of lungs, the mouth cavity, etc.
                                  \n2. The cells are nearly hexagonal, flat and plate-like.
                                  \n\"1st
                                  \n3. A spherical or oval nucleus is found at the cell centre.
                                  \n4. The cells are closely fitted without intercellular space. Hence look like a mosaic pavement. Because of this squamous epithelium is also called pavement epithelium.<\/p>\n

                                  Function:
                                  \n1. It forms a continuous cover and protects the inner tissues.
                                  \n2. It helps in ultrafiltration in Bowman\u2019s capsule.
                                  \n3. It helps in the exchange of gases in alveoli.<\/p>\n

                                  Note: Squamous epithelium lining the blood vessels is called endothelium( derived from endoderm). Squamous epithelium lining internal organs such as the stomach, the intestine is called mesothelium (derived from mesoderm).<\/p>\n

                                  Question 35.
                                  \nExplain the pressure-flow hypothesis of the translocation of organic solutes.
                                  \nAnswer:
                                  \nAccording to Munch, food materials are translocated from the source (leaves) to sink (roots) enmass, through a turgor pressure gradient that occurs between leaves and roots. The mechanism of translocation can be explained as follows:<\/p>\n

                                  1. Sugars are fonned in the mesophyll cells during photosynthesis and are loaded into the phloem sieve elements of veins. This is called vein loading.<\/p>\n

                                  2. The osmotic potential in the phloem becomes more negative, and as a result, water is drawn into phloem elements from xylem cells.<\/p>\n

                                  3. Turgor pressure increases in the sieve tubes of leaves, and at the same time the turgor. the pressure of roots becomes less. So a turgor pressures gradient occurs between mesophyll cells and root cells.<\/p>\n

                                  4. As a result, food molecules in solution form, are translocated from leaves Tc roots.<\/p>\n

                                  5. In the roots the food \u00a1s consumed, and in the fruits, it is converted into insoluble starch. So osmotic potential and turgor pressure in the roots and storage organs decreases.<\/p>\n

                                  6. Thus turgor pressure gradient continues between leaves, and roots and translocation of food also continue.
                                  \n\"1st
                                  \nMunch conducted a physical experiment to demonstrate the mass flow. He took two osmometers A and B, with concentrated sugar solution in A and B filled with a dilute solution.<\/p>\n

                                  These are connected by a \u2018U\u2019 tube C. A B is immersed in a trough containing water.<\/p>\n

                                  Now water enters into \u2018A\u2019, due to high osmotic pressure and this creates turgor pressure.<\/p>\n

                                  This results in the mass flow of sugar solution from A to B through \u2018C\u2019 until equilibrium is attained and maintained. He compared osmometer A to leaves, B to roots, C to phloem, and water in vessels to xylem vessels. This clearly explains the mass flow hypothesis.<\/p>\n

                                  \"KSEEB<\/p>\n

                                  Question 36.
                                  \n(a) Explain the mechanism of clotting of blood. (3M)
                                  \n(b) Differentiate pulmonary and systemic circulations. (2M)
                                  \nAnswer:
                                  \nBlood has a unique property. As long as circulating inside the body it retains the fluid state however as soon as blood comes out of the body as a result of a cut or an injury, blood is transformed from fluid state to gel state, preventing further loss. This self-regulating mechanism of the blood is known as clotting of the blood.<\/p>\n

                                  The mechanism of blood clotting is explained by various theories out of these \u2018Best and Taylor\u2019s theory is one. According to this theory. four factors are responsible for blood clotting.<\/p>\n

                                    \n
                                  1. Prothrombin: Produced by the liver and present in plasma.<\/li>\n
                                  2. Thmmboplastin: An enzyme released from damaged tissues, present in the body tissues.<\/li>\n
                                  3. Calcium ions: Present in the plasma of the blood.<\/li>\n
                                  4. Fibrinogen: Produced by the liver and present in plasma.<\/li>\n<\/ol>\n

                                    Chemical events leading to the formation of b od clot as suggested by \u2018Best and Taylor’s theory can be represented as follows.
                                    \n\"1st
                                    \nThe reaction shown above indicates that the inactive Prothrombin is converted into active thrombin by the catalytic activity of the enzyme Thromboplastin in the presence of Ca2<\/sub>. The activated thrombin reacts with soluble fibrinogen, resulting in the formation of insoluble fibrin.<\/p>\n

                                    The fibrin threads form a mesh-like network in which the blood components get entangled, and results in blood clots and thus prevents bleeding.<\/p>\n

                                    Best and Taylor\u2019s theory doesn\u2019t consider the role of platelets in the blood clotting mechanism.<\/p>\n

                                    Question 37.
                                    \nDraw a labelled diagram of the human brain in the sagittal section.
                                    \nAnswer:
                                    \n\"1st<\/p>\n","protected":false},"excerpt":{"rendered":"

                                    Students can Download 1st PUC Biology Model Question Paper 9 with Answers, Karnataka 1st PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Biology Model Question Paper 9 with Answers Time: 3.15 Hours Max Marks: 70 General …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89777"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=89777"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/89777\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=89777"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=89777"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=89777"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}